# Thread: Chain rule question...

1. ## Chain rule question...

Given the function:
$s(t)=\sin2\pi*t$

My book says: $s'(t)=2\pi*\cos2\pi*t$

I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

(ps, sorry I could not get it to show up any clearer...)

2. chain rule is g'(h(x)) h'(x) . which means..

g(x)=sin(x) [x being sin(2Pi)]
g'(x)=cos(x)
h(x)=sin(2Pi) and
h'(x)=0 .

put it all into the function and you get..

cos(sin(2Pi)) (0)

which ends up giving you a 0 anyways. are you sure thats what the book is saying? lol i mean it sounds dumb to ask but this is the way i did it.. idk and i tried many times on the calculator and i keep getting zero because the derivative of t is 1.. and anything that you multiply by zero stays zero... you know??? any more info you have?

3. Originally Posted by rust1477
Given the function:
$s(t)=\sin2\pi*t$

My book says: $s'(t)=2\pi*\cos2\pi*t$

I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

(ps, sorry I could not get it to show up any clearer...)
Is s(t) = sin(2pi times t), or

sin(2pi) times t?

makes it easier for me to help you hehe

4. yea! lol.. im pretty sure thats it.. that makes more sense. hehe how come i didnt see that?? good job.

5. Originally Posted by rust1477
Given the function:
$s(t)=\sin2\pi*t$

My book says: $s'(t)=2\pi*\cos2\pi*t$

I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

(ps, sorry I could not get it to show up any clearer...)
Clearly the function is $s(t) = \sin (2 \pi t)$. Please post your equations so that they are less ambiguous.

You're expected to know the standard form: If $y = \sin (kt)$ then $\frac{dy}{dt} = k \cos (kt)$.